Question

What is the `"pH"` for an aqueous solution with a hydroxide concentration of `5 xx 10^(-3) "M"` at `25^@ "C"`?

Answer 1

`11.7`

Explanation:

`"pOH" = - log["OH"^-]`

`"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30`

Now,

`"pH" + "pOH" = 14`

`"pH" = 14 - "pOH" = 14 - 2.30 = 11.7`

Answer 2

`"pH" ~~ color(blue)(11.7)`

Explanation:

`color (indigo)(pH=-log [H_3O^+])`
`color (indigo)(pOH=-log [OH^-])`
`color(cyan)([OH^-]=0.005)`
Here,
`pOH=-log [0.005]`
`pOH=-log [5×10^-3]`
`pOH=2.3010299957`

`color (green)(pH=14-pOH)`

`ph=14-2.301 =11.699` `~~` `bb11.7`